- #1
hbweb500
- 40
- 1
So I am playing around with the differential form of Gauss's Law:
\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}
Starting off simple with a point charge, the electric field is:
\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r}
And the divergence, in spherical coordinates, is:
\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r)
= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \right)
= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{q}{4 \pi \epsilon_0}} \right)
= 0
I can handle this much. It makes sense that the divergence is everywhere 0, since the only charge density is a point charge. Griffiths has a discussion about this very thing, where he states that the infinite divergence at the origin causes things to work out as expected.
My problem is in dealing with a cartesian coordinate system. I didn't recall the divergence equation in spherical coordinates when I was first playing around with this, so I tried it in cartesian. Here:
\vec{E} = \frac{q}{4\pi \epsilon_0} \frac{ x \hat{x} + y \hat{y} }{(x^2+y^2)^{\frac{3}{2}}}
But the divergence of this is proportional to:
\nabla \cdot \vec{E} \propto \frac{-2}{(x^2 + y^2)^3}
Which clearly isn't zero everywhere. I've checked my divergence and electric field equation, but I can't find the difference between it and the spherical ones I am using.
So, what gives? Is the divergence in spherical coordinates 0, but nonzero in cartesian?
\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}
Starting off simple with a point charge, the electric field is:
\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r}
And the divergence, in spherical coordinates, is:
\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r)
= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \right)
= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{q}{4 \pi \epsilon_0}} \right)
= 0
I can handle this much. It makes sense that the divergence is everywhere 0, since the only charge density is a point charge. Griffiths has a discussion about this very thing, where he states that the infinite divergence at the origin causes things to work out as expected.
My problem is in dealing with a cartesian coordinate system. I didn't recall the divergence equation in spherical coordinates when I was first playing around with this, so I tried it in cartesian. Here:
\vec{E} = \frac{q}{4\pi \epsilon_0} \frac{ x \hat{x} + y \hat{y} }{(x^2+y^2)^{\frac{3}{2}}}
But the divergence of this is proportional to:
\nabla \cdot \vec{E} \propto \frac{-2}{(x^2 + y^2)^3}
Which clearly isn't zero everywhere. I've checked my divergence and electric field equation, but I can't find the difference between it and the spherical ones I am using.
So, what gives? Is the divergence in spherical coordinates 0, but nonzero in cartesian?
